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Question Number 183324 by Acem last updated on 25/Dec/22
Find the minimum distance between C_f , C_g ; C_f : y^2 = 4ax , C_g : x^2 + y^2 −24ay+ 128a^2 = 0
$${Find}\:{the}\:{minimum}\:{distance}\:{between}\:{C}_{{f}} \:,\:{C}_{{g}} \\ $$$$\:;\:{C}_{{f}} \::\:{y}^{\mathrm{2}} =\:\mathrm{4}{ax}\:,\:{C}_{{g}} :\:{x}^{\mathrm{2}} +\:{y}^{\mathrm{2}} −\mathrm{24}{ay}+\:\mathrm{128}{a}^{\mathrm{2}} =\:\mathrm{0} \\ $$
Commented by Acem last updated on 25/Dec/22
No matter if ∀a ∈ R^(∗+) or R^(∗−)
$$\:{No}\:{matter}\:{if}\:\:\forall{a}\:\in\:\mathbb{R}^{\ast+} \:{or}\:\mathbb{R}^{\ast−} \\ $$
Answered by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
a≠0, assume a>0 y^2 =4ax x^2 +(y−12a)^2 =(4a)^2 A(0,12a) r=4a say P((p^2 /(4a)), p) tan θ=(dy/dx)=((2a)/y)=((2a)/p)=((p^2 /(4a))/(12a−p)) ⇒((p/a))^3 +8((p/a))−96=0 ⇒(p/a)=4 R=AP=(√(((p^2 /(4a)))^2 +(12a−p)^2 ))=4(√5)∣a∣ d_(min) =R−r=4((√5)−1)∣a∣
$${a}\neq\mathrm{0},\:{assume}\:{a}>\mathrm{0} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax} \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{12}{a}\right)^{\mathrm{2}} =\left(\mathrm{4}{a}\right)^{\mathrm{2}} \\ $$$${A}\left(\mathrm{0},\mathrm{12}{a}\right) \\ $$$${r}=\mathrm{4}{a} \\ $$$${say}\:{P}\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}{a}},\:{p}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\frac{\mathrm{2}{a}}{{y}}=\frac{\mathrm{2}{a}}{{p}}=\frac{\frac{{p}^{\mathrm{2}} }{\mathrm{4}{a}}}{\mathrm{12}{a}−{p}} \\ $$$$\Rightarrow\left(\frac{{p}}{{a}}\right)^{\mathrm{3}} +\mathrm{8}\left(\frac{{p}}{{a}}\right)−\mathrm{96}=\mathrm{0} \\ $$$$\Rightarrow\frac{{p}}{{a}}=\mathrm{4} \\ $$$${R}={AP}=\sqrt{\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}{a}}\right)^{\mathrm{2}} +\left(\mathrm{12}{a}−{p}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{5}}\mid{a}\mid \\ $$$${d}_{{min}} ={R}−{r}=\mathrm{4}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\mid{a}\mid \\ $$
Commented by Acem last updated on 25/Dec/22
Thx Sir!
$${Thx}\:{Sir}! \\ $$
Answered by Acem last updated on 25/Dec/22
For a> 0 C_(circle) (0, 12a) , R= 4a y= 2(√(a ))(√x) ^(The upper part of the parabola) Assume that x= ak^2 then y= 2ak , P(ak^2 , 2ak)∈ C_f The normal △ passes through P and the center of the circle C △: y+ xk = 2ak+ ak^3 = 12a_(x_c = 0) ⇔ k^3 +2k−12=0 ⇒ (k−2)(k^2 +2k+6)_(No solution in R) = 0 ⇒ k= 2 ⇒P(4a, 4a) , T∈ C_g , △ ∣TP∣_(min. Dis.) = (√((−4a)^2 +(12a−4a)^2 )) − 4a= 4a((√5)−1) uni. ; ∣TP∣= ∣CP∣ − R
$$\:{For}\:{a}>\:\mathrm{0} \\ $$$$\:{C}_{{circle}} \left(\mathrm{0},\:\mathrm{12}{a}\right)\:,\:{R}=\:\mathrm{4}{a} \\ $$$$\:{y}=\:\mathrm{2}\sqrt{{a}\:}\sqrt{{x}}\:\:^{{The}\:{upper}\:{part}\:{of}\:{the}\:{parabola}} \\ $$$$\:{Assume}\:{that}\:{x}=\:{ak}^{\mathrm{2}} \:{then}\:{y}=\:\mathrm{2}{ak}\:\:,\:{P}\left({ak}^{\mathrm{2}} ,\:\mathrm{2}{ak}\right)\in\:{C}_{{f}} \\ $$$$\:{The}\:{normal}\:\bigtriangleup\:{passes}\:{through}\:{P}\:{and}\:{the}\:{center}\:{of}\:{the}\:{circle}\:{C} \\ $$$$\bigtriangleup:\:{y}+\:{xk}\:=\:\mathrm{2}{ak}+\:{ak}^{\mathrm{3}} =\:\mathrm{12}{a}_{{x}_{{c}} =\:\mathrm{0}} \:\Leftrightarrow\:{k}^{\mathrm{3}} +\mathrm{2}{k}−\mathrm{12}=\mathrm{0} \\ $$$$\:\Rightarrow\:\left({k}−\mathrm{2}\right)\underset{{No}\:{solution}\:{in}\:\mathbb{R}} {\underbrace{\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{6}\right)}}=\:\mathrm{0}\:\Rightarrow\:{k}=\:\mathrm{2} \\ $$$$\:\Rightarrow{P}\left(\mathrm{4}{a},\:\mathrm{4}{a}\right)\:,\:{T}\in\:{C}_{{g}} \:,\:\bigtriangleup \\ $$$$ \\ $$$$\:\mid{TP}\mid_{{min}.\:{Dis}.} =\:\sqrt{\left(−\mathrm{4}{a}\right)^{\mathrm{2}} +\left(\mathrm{12}{a}−\mathrm{4}{a}\right)^{\mathrm{2}} }\:−\:\mathrm{4}{a}=\:\mathrm{4}{a}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\:{uni}. \\ $$$$\:;\:\mid{TP}\mid=\:\mid{CP}\mid\:−\:{R} \\ $$$$ \\ $$

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